Is tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) ?

The answer to whether tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) is True

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prove tan^2(x)-sin^2(x)=sin^4(x)sec^2(x)

prove

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

True

Solution steps

tan^{2} (x) - sin^{2} (x) = sin^{4} (x) sec^{2} (x)

Manipulating left side

tan^{2} (x) - sin^{2} (x)

Express with sin, cos

= \frac{sin ^{2} ( x ) - cos ^{2} ( x ) sin ^{2} ( x )}{cos ^{2} ( x )}

Factor

\frac{sin ^{2} ( x ) - cos ^{2} ( x ) sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{sin ^{2} ( x ) ( 1 - cos ^{2} ( x ) )}{cos ^{2} ( x )}

= \frac{( 1 - cos ^{2} ( x ) ) sin ^{2} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

= \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

sin^{4} (x) sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (x) (sec^{2} (x) - 1) = sin^{2} (x)

\frac{csc ( x ) - cot ( x )}{sec ( x ) - 1} = cot (x)

\frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

\frac{1 - tan ^{2} ( x )}{cot ^{2} ( x ) - 1} = tan^{2} (x)

\frac{2 tan ( θ )}{1 + tan ^{2} ( θ )} = sin (2 θ)

Is tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) ?
The answer to whether tan^2(x)-sin^2(x)=sin^4(x)sec^2(x) is True